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Filling the Complexity Gaps for Colouring Planar and Bounded Degree Graphs

06/22/2015
by   Konrad K. Dabrowski, et al.
ENS Lyon
Durham University
0

We consider a natural restriction of the List Colouring problem: k-Regular List Colouring, which corresponds to the List Colouring problem where every list has size exactly k. We give a complete classification of the complexity of k-Regular List Colouring restricted to planar graphs, planar bipartite graphs, planar triangle-free graphs and to planar graphs with no 4-cycles and no 5-cycles. We also give a complete classification of the complexity of this problem and a number of related colouring problems for graphs with bounded maximum degree.

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1 Introduction

A colouring of a graph is a labelling of the vertices so that adjacent vertices do not have the same label. We call these labels colours. Graph colouring problems are central to the study of combinatorial algorithms and they have many theoretical and practical applications. A typical problem asks whether a colouring exists under certain constraints, or how difficult it is to find such a colouring. For example, in the List Colouring problem, a graph is given where each vertex has a list of colours and one wants to know if the vertices can be coloured using only colours in their lists. The Choosability problem asks whether such list colourings are guaranteed to exist whenever all the lists have a certain size. In fact, an enormous variety of colouring problems can be defined and there is now a vast collection of literature on this subject. For longer introductions to the type of problems we consider we refer to two recent surveys [9, 16].

In this paper, we are concerned with the computational complexity of colouring problems. For many such problems, the complexity is well understood in the case where we allow every graph as input, so it is natural to consider problems with restricted inputs. We consider a variant of the List Colouring problem, closely related to Choosability, and give a complete classification of its complexity for planar graphs and a number of subclasses of planar graphs by combining known results with new results. Some of the known results are for (planar) graphs with bounded degree. We use these results to fill some more complexity gaps by giving a complete complexity classification of a number of colouring problems for graphs with bounded maximum degree.

1.1 Terminology

A colouring of a graph is a function such that whenever . We say that  is the colour of . For a positive integer , if for all then  is a -colouring of . We say that  is -colourable if a -colouring of  exists. The Colouring problem is to decide whether a graph  is -colourable for some given integer . If  is fixed, that is, not part of the input, we obtain the -Colouring problem.

A list assignment of a graph is a function  with domain  such that for each vertex , is a subset of . This set is called the list of admissible colours for . If for each , then  is a -list assignment. The size of a list assignment  is the maximum list size  over all vertices . A colouring  respects  if for all . Given a graph  with a -list assignment , the List Colouring problem is to decide whether  has a colouring that respects . If  is fixed, then we have the List -Colouring problem. Fixing the size of  to be at most  gives the -List Colouring problem. We say that a list assignment  of a graph is -regular if, for all , contains exactly  colours. This gives us the following problem, which is one focus of this paper. It is defined for each integer (note that  is fixed; that is, is not part of the input).

-Regular List Colouring
Instance:
a graph  with an -regular list assignment .
Question: does  have a colouring that respects ?

A -precolouring of a graph is a function for some subset . A -colouring  of  is an extension of a -precolouring  of  if for each . Given a graph  with a precolouring , the Precolouring Extension problem is to decide whether  has a -colouring that extends . If  is fixed, we obtain the -Precolouring Extension problem.

The relationships amongst the problems introduced are shown in Figure 1.

-Regular List Colouring

-List Colouring

-List Colouring

List Colouring

List -Colouring

List -Colouring

-Precolouring Extension

-Colouring

Precolouring Extension

Colouring
Figure 1: Relationships between Colouring and its variants. An arrow from one problem to another indicates that the latter is (equivalent to) a special case of the former; and  are any two arbitrary integers for which . For instance, -Colouring is a special case of -Regular List Colouring. This can be seen by giving the list to each vertex  in an instance graph of Colouring. We also observe that -Regular List Colouring and -Regular List Colouring are not comparable for any .

For an integer , a graph is -choosable if, for every -regular list assignment  of , there exists a colouring that respects . The corresponding decision problem is the Choosability problem. If  is fixed, we obtain the -Choosability problem.

We emphasize that -Regular List Colouring and -Choosability are two fundamentally different problems. For the former we must decide whether there exists a colouring that respects a particular -regular list assignment. For the latter we must decide whether or not every -regular list assignment has a colouring that respects it. As we will see later, this difference also becomes clear from a complexity point of view: for some graph classes -Regular List Colouring is computationally easier than -Choosability, whereas, perhaps more surprisingly, for other graph classes, the reverse holds.

For two vertex-disjoint graphs  and , we let denote the disjoint union , and  denote the disjoint union of  copies of . If  is a graph containing an edge  or a vertex  then and denote the graphs obtained from  by deleting  or , respectively. If  is a subgraph of  then denotes the graph with vertex set  and edge set . We let , and  denote the cycle, complete graph and path on  vertices, respectively. A wheel is a cycle with an extra vertex added that is adjacent to all other vertices. The wheel on  vertices is denoted ; note that . A graph on at least three vertices is -connected if it is connected and there is no vertex whose removal disconnects it. A block of a graph is a maximal subgraph that is connected and cannot be disconnected by the removal of one vertex (so a block is either -connected a  or an isolated vertex). A block graph is a connected graph in which every block is a complete graph. A Gallai tree is a connected graph in which every block is a complete graph or a cycle. We say that  is a leaf-block of a connected graph  if  contains exactly one cut vertex  of  and is a component of . For a set of graphs , a graph is -free if contains no induced subgraph isomorphic to a graph in , whereas is -subgraph-free if it contains no subgraph isomorphic to a graph in . The girth of a graph is the length of its shortest cycle.

1.2 Known Results for Planar Graphs

We start with a classical result observed by Erdős et al. [13] and Vizing [27].

Theorem 1.1 ([13, 27])

-List Colouring is polynomial-time solvable.

Garey et al. proved the following result, which is in contrast to the fact that every planar graph is -colourable by the Four Colour Theorem [2].

Theorem 1.2 ([14])

-Colouring is NP-complete for planar graphs of maximum degree .

Next we present results found by several authors on the existence of -choosable graphs for various graph classes.

Theorem 1.3

The following statements hold for -choosability:

  1. Every planar graph is -choosable [25].

  2. There exists a planar graph that is not -choosable [29].

  3. Every planar triangle-free graph is -choosable [20].

  4. Every planar graph with no -cycles is -choosable [21].

  5. There exists a planar triangle-free graph that is not -choosable [30].

  6. There exists a planar graph with no -cycles, no -cycles and no intersecting triangles that is not -choosable [24].

  7. Every planar bipartite graph is -choosable [1].

We note that smaller examples of graphs than were used in the original proofs have been found for Theorems 1.3.(ii) [18], 1.3.(v) [23] and 1.3.(vi) [34] and that Theorem 1.3.(vi) strengthens a result of Voigt [31]. We recall that Thomassen [26] first showed that every planar graph of girth at least  is -choosable, and that a number of results have since been obtained on -choosability of planar graphs in which certain cycles are forbidden; see, for example, [7, 11, 32, 33].

We will also use the following result of Chlebík and Chlebíková.

Theorem 1.4 ([8])

List Colouring is NP-complete for -regular planar bipartite graphs that have a list assignment in which each list is one of , , , and all the neighbours of each vertex with three colours in its list have two colours in their lists.

1.3 New Results for Planar Graphs

Theorems 1.11.3 have a number of immediate consequences for the complexity of -Regular List Colouring when restricted to planar graphs. For instance, Theorem 1.2 implies that 3-Regular List Colouring is NP-complete for planar graphs, whereas Theorem 1.3.(i) shows that 5-Regular List Colouring is polynomial-time solvable on this graph class. As such, it is a natural question to determine the complexity for the missing case . In this section we settle this missing case and also present a number of new hardness results for -Regular List Colouring restricted to various subclasses of planar graphs. At the end of this section we show how to combine the known results with our new ones to obtain a number of dichotomy results (Corollaries 36). We deduce some of our new results from two more general theorems, namely Theorems 1.5 and 1.6, which we state below, but which we prove in Sections 2.1 and 2.2, respectively.

Theorem 1.5

Let  be a finite set of -connected planar graphs. Then -Regular List Colouring is NP-complete for planar -subgraph-free graphs if there exists a planar -subgraph-free graph that is not -choosable.

Note that the class of -subgraph-free graphs is contained in the class of -free graphs. Hence, whenever a problem is NP-complete for -subgraph-free graphs, it is also NP-complete for -free graphs.

Combining Theorem 1.5 with Theorem 1.3.(ii) yields the following result which, as we will see later, was the only case for which the complexity of -Regular List Colouring for planar graphs was not settled.

Corollary 1

-Regular List Colouring is NP-complete for planar graphs.

Theorem 1.5 has more applications. For instance, consider the non-4-choosable planar graph  from the proof of Theorem 1.7 in [18]. It can be observed that  is -subgraph-free for all . Wheels are -connected and planar. Hence if  is any finite set of wheels on at least eight vertices then -Regular List Colouring is NP-complete for planar -subgraph-free graphs.

Our basic idea for proving Theorem 1.5 is to pick a minimal counterexample  with list assignment  (which may exist due to Theorem 1.3.(ii)). We select an “appropriate” edge and consider the graph . We reduce from an appropriate colouring problem restricted to planar graphs and use copies of  as a gadget to ensure that we can enforce a regular list assignment. The proof of the next theorem also uses this idea.

Theorem 1.6

Let  be a finite set of -connected planar graphs. Then -Regular List Colouring is NP-complete for planar -subgraph-free graphs if there exists a planar -subgraph-free graph that is not -choosable.

Theorem 1.6 has a number of applications. For instance, if we let then Theorem 1.6, combined with Theorem 1.3.(v), leads to the following result.

Corollary 2

-Regular List Colouring is NP-complete for planar triangle-free graphs.

Theorem 1.6 can also be used for other classes of graphs. For example, let  be a finite set of graphs, each of which includes a -connected graph on at least five vertices as a subgraph. Let  be the set of these -connected graphs. The graph  is a planar -subgraph-free graph that is not -choosable (since it is not -colourable). Therefore, Theorem 1.6 implies that -Regular List Colouring is NP-complete for planar -subgraph-free graphs. We can obtain more hardness results by taking some other planar graph that is not -choosable, such as a wheel on an even number of vertices. Also, if we let we can use Theorem 1.6 by combining it with Theorem 1.3.(vi) to find that -Regular List Colouring is NP-complete for planar graphs with no -cycles and no -cycles. We strengthen this result as follows (see Section 2.3 for the proof).

Theorem 1.7

-Regular List Colouring is NP-complete for planar graphs with no -cycles, no -cycles and no intersecting triangles.

Corollaries 1 and 2 and Theorem 1.7 can be seen as strengthenings of Theorems 1.3.(ii)1.3.(v) and 1.3.(vi), respectively. Moreover, they complement Theorem 1.2, which implies that -List Colouring is NP-complete for planar graphs, and a result of Kratochvíl [19] that, for planar bipartite graphs, -Precolouring Extension is NP-complete. Corollaries 1 and 2 also complement results of Gutner [18] who showed that -Choosability and -Choosability are -complete for planar triangle-free graphs and planar graphs, respectively. However, we emphasize that, for special graph classes, it is not necessarily the case that -Choosability is computationally harder than -Regular List Colouring. For instance, contrast the fact that Choosability is polynomial-time solvable on -free graphs [15] with our next result, which we prove in Section 2.4.

Theorem 1.8

-Regular List Colouring is NP-complete for -free graphs.

Our new results, combined with known results, close a number of complexity gaps for the -Regular List Colouring problem. Combining Corollary 1 with Theorems 1.11.2 and 1.3.(i) gives us Corollary 3. Combining Theorem 1.7 with Theorems 1.1 and 1.3.(iv) gives us Corollary 4. Combining Corollary 2 with Theorems 1.1 and 1.3.(iii) gives us Corollary 5, whereas Theorems 1.1 and 1.3.(vii) imply Corollary 6.

Corollary 3

Let  be a positive integer. Then -Regular List Colouring, restricted to planar graphs, is NP-complete if and polynomial-time solvable otherwise.

Corollary 4

Let  be a positive integer. Then -Regular List Colouring, restricted to planar graphs with no -cycles and no -cycles and no intersecting triangles, is NP-complete if and polynomial-time solvable otherwise (even if we allow intersecting triangles and -cycles).

Corollary 5

Let  be a positive integer. Then -Regular List Colouring, restricted to planar triangle-free graphs, is NP-complete if and polynomial-time solvable otherwise.

Corollary 6

Let  be a positive integer. Then -Regular List Colouring, restricted to planar bipartite graphs, is polynomial-time solvable.

1.4 Known Results for Bounded Degree Graphs

First we present a result of Kratochvíl and Tuza [20].

Theorem 1.9 ([20])

List Colouring is polynomial-time solvable on graphs of maximum degree at most .

Brooks’ Theorem [6] states that every graph  with maximum degree  has a -colouring unless 

is a complete graph or a cycle with an odd number of vertices. The next result of Vizing 

[28] generalizes Brooks’ Theorem to list colourings.

Theorem 1.10 ([28])

Let  be a positive integer. Let be a connected graph of maximum degree at most  and let  be a -regular list assignment for . If  is not a cycle or a complete graph then  has a colouring that respects .

And we need another result of Chlebík and Chlebíková [8].

Theorem 1.11 ([8])

Precolouring Extension is polynomial-time solvable on graphs of maximum degree .

1.5 New Results for Bounded Degree Graphs

In Section 2.5, we prove the following result by making a connection to Gallai trees.

Theorem 1.12

Let  be a positive integer. Then -Precolouring Extension is polynomial-time solvable for graphs of maximum degree at most .

We have the following two classifications. The first one is an observation obtained by combining only previously known results, whereas the second one also makes use of our new result.

Corollary 7

Let  be a positive integer. The following two statements hold for graphs of maximum degree at most .

  1. List Colouring is NP-complete if and polynomial-time solvable if .

  2. Precolouring Extension and Colouring are NP-complete if and polynomial-time solvable if .

Proof

We first consider (i). If , we use Theorem 1.4. If , we use Theorem 1.9. We now consider (ii). If , we use Theorem 1.2. If , we use Theorem 1.11.∎

Corollary 8

Let  and  be two positive integers. The following two statements hold for graphs of maximum degree at most .

  1. -List Colouring and List -Colouring are NP-complete if and and polynomial-time solvable otherwise.

  2. -Regular List Colouring and -Precolouring Extension are NP-complete if and and polynomial-time solvable otherwise.

Proof

We first consider (i). If and , we use Theorem 1.4. If or , we use Theorems 1.1 or 1.9, respectively.

We now consider (ii). We start with the hardness cases and so let and .

First consider -Precolouring Extension. Theorem 1.2 implies that -Colouring is NP-complete for graphs of maximum degree at most  for all . The case follows immediately from this result. Suppose and . Consider a graph  of maximum degree . For each vertex , we add new vertices and edges . Let  be the resulting graph. Note that  has maximum degree at most . We define a precolouring  on the newly added vertices by assigning colour to each . Then  has a -colouring extending  if and only if  has a -colouring.

Now consider -Regular List Colouring. The case follows immediately from Theorem 1.2. Suppose and . Consider a graph  of maximum degree . We define the list for each vertex . For each vertex , we add new vertices and edges . We define the list for each . For each vertex , we also add  new vertices and edges such that form a clique (on vertices). We define the list for each . Let  be the resulting graph. Note that  has maximum degree at most and that the resulting list assignment  is a -regular list assignment of . Then  has a -colouring respecting  if and only if  has a -colouring.

We continue with the polynomial-time solvable cases. If , the result follows from Theorem 1.1. Suppose that and . Then the result for -Regular List Colouring follows from Theorems 1.9 and 1.10 and the result for -Precolouring Extension follows from Theorem 1.12.∎

Note that Corollary 8 does not contain a dichotomy for -Colouring restricted to graphs of maximum degree at most . A full classification of this problem is open, but a number of results are known. Molloy and Reed [22]classified the complexity for all pairs for sufficiently large . Emden-Weinert et al. [12] proved that -Colouring is NP-complete for graphs of maximum degree at most .

2 Proofs

2.1 The Proof of Theorem 1.5

We need an additional result.

Theorem 2.1

For every integer , -List Colouring is NP-complete for planar graphs of girth at least  that have a list assignment in which each list is one of , , , .

Proof

By Theorem 1.4, List Colouring is NP-complete for -regular planar bipartite graphs that have a list assignment in which each list is one of , , , and all the neighbours of each vertex with three colours in its list have two colours in their lists. We modify the hardness construction as follows. Note that for each edge at least one of the incident vertices has a list of size . We replace each edge by a path on an odd number of edges in such a way that the girth of the graph obtained is at least . The new vertices on the path are all given the same list of size , identical to the list on one or other of the end-vertices. It is readily seen that these modifications do not affect whether or not the graph can be coloured.∎

We are now ready to prove Theorem 1.5, which we restate below.

Theorem 1.5 (restated). Let  be a finite set of -connected planar graphs. Then -Regular List Colouring is NP-complete for planar -subgraph-free graphs if there exists a planar -subgraph-free graph that is not -choosable.

Proof

The problem is readily seen to be in NP. Let  be a planar -subgraph-free graph with a -regular list assignment  such that  has no colouring respecting . We may assume that  is minimal (with respect to the subgraph relation). In particular, this means that  is connected. Let  be the length of a longest cycle in any graph of . We reduce from the problem of -List Colouring restricted to planar graphs of girth at least in which each vertex has list , , or . This problem is NP-complete by Theorem 2.1. Let a graph  and list assignment  be an instance of this problem. We will construct a planar -subgraph-free graph  with a -regular list assignment  such that  has a colouring that respects  if and only if  has a colouring that respects .

If every pair of adjacent vertices in  has the same list, then the problem of finding a colouring that respects  is just the problem of finding a -colouring which, by the Four Colour Theorem [2], we know is possible. Thus we may assume that, on the contrary, there is an edge such that . Let . Then, by minimality, has at least one colouring respecting , and moreover, for any colouring of  that respects , and  are coloured alike (otherwise we would have a colouring of  that respects ). Let  be the set of possible colours that can be used on  and  in colourings of  that respect  and let . As , we have . Up to renaming the colours in , we can build copies of  with -regular list assignments such that

  1. the set is any given list of colours of size , and

  2. the vertex corresponding to  has any given list of  colours containing .

We will implicitly make use of this several times in the remainder of the proof.

We say that a vertex  in  is a bivertex or trivertex if  is  or , respectively. We construct a planar -subgraph-free graph  and -regular list assignment  as follows.

First suppose that . For each bivertex  in , we do as follows. We add two copies of  to , which we label  and . The vertex in  corresponding to  is labelled  for and we set . We add the edges  and . We give list assignments to the vertices of  and  such that for  and for . We let . For each trivertex  in , we do as follows. We add one copy of  to , which we label . The vertex in  corresponding to  is labelled  and we set . We add the edge . We give list assignments to vertices of such that for . We let . This completes the construction of  and  when .

Now suppose that . Let if  is even and if  is odd (so  is even in both cases). For each bivertex  in , we do as follows. We add a copy of  to , which we label , and identify the vertex in  corresponding to  with . We give list assignments to vertices of such that and . For each trivertex  in , we do as follows. We add  copies of  to  which we label , . The vertex in  corresponding to  is labelled . Let . Add edges such that the union of  and  induces a cycle on vertices. For all , we give list assignments to vertices of  such that . We let . This completes the construction of  and  when .

Now suppose that . For each bivertex  in , we do as follows. We add two copies of  to  which we label  and , such that for , the vertex in  corresponding to  is identified with . We give list assignments to vertices of  and  such that for , for  and . For each trivertex  in , we do as follows. We add a copy of  to  which we label , such that the vertex in  corresponding to  is identified with . We give list assignments to the vertices of  such that and . This completes the construction of  and  when .

Note that  is planar. Suppose that there is a subgraph  in  that is isomorphic to a graph of . Since  is -subgraph-free, and since  is obtained from  by removing one edge, is also -subgraph-free. Therefore for all , is not fully contained in any . Since  is -connected and since for all  only one vertex of any  has a neighbour outside of , we find that  has at most one vertex in each . In particular, cannot contain any vertex of any  in which the vertex corresponding to  has been attached to  (as opposed to being identified with ); this includes the case when the union of  and  induces a cycle on vertices. Hence we have found that  is a subgraph of , which contradicts the fact that  has girth at least . Therefore  is -subgraph-free.

Note that in any colouring of  that respects , each copy of  must be coloured such that the vertices corresponding to  and  have the same colour, which must be one of the colours from the corresponding set . If and  is a trivertex, this means that the unique neighbour of  in  must be coloured with colour , so  cannot be coloured with colour . Similarly, if and  is a bivertex or and  is a trivertex then the two neighbours of  in  must be coloured with colours  and , so  cannot be coloured with colours  or . If and  is a bivertex or and  is a trivertex then  belongs to a copy of  with , so  cannot have colour  or . If and  is a bivertex then  belongs to two copies of , one with and one with . Therefore, must be coloured with a colour from the intersection of these two sets, that is it must be coloured with a colour from . Therefore none of the vertices of  can be coloured  or . Thus the problem of finding a colouring of  that respects  is equivalent to the problem of finding a colouring of  that respects . This completes the proof.∎

2.2 The Proof of Theorem 1.6

Theorem 1.6 (restated). Let  be a finite set of -connected planar graphs. Then -Regular List Colouring is NP-complete for planar -subgraph-free graphs if there exists a planar -subgraph-free graph that is not -choosable.

Proof

The problem is readily seen to be in NP. Every graph in  is -connected, and therefore contains a cycle. Let  be the length of a longest cycle in any graph of . By assumption, there exists a planar graph  and -regular list assignment  such that  is -subgraph-free and has no colouring respecting . We may assume that  is minimal by removing edges and vertices until any further removal would give a graph with a colouring respecting . In particular, we note that  is connected.

We distinguish two cases.

Case 1.  is the same for every vertex  in .
Then we may assume without loss of generality that for all . We reduce from -Colouring which is NP-complete even for planar graphs by Theorem 1.2. Let  be a planar graph. We will construct a planar -subgraph-free graph  as follows.

Let be an edge of . Let . Then, by minimality, has at least one colouring respecting , which must be a -colouring as every list consists of the colours . For every -colouring  of , it holds that (otherwise  would be a colouring of  that respects ). Moreover, since we can permute the colours, there is such a colouring  that colours  (and thus ) with colour  for each . Note that in  the vertices  and  must be at distance at least  from each other.

Let . Assume that the vertices of  are ordered. For each edge with , we do the following:

  1. delete ;

  2. add  copies of  labelled and, for , let  and  be the vertices in  corresponding to  and ;

  3. identify  with  and, for , identify  with ;

  4. add an edge from  to .

Let  be the obtained graph and note that  is planar. Every cycle in  that is not contained in a copy of  has length at least , since it corresponds to a cycle in  and in which every edge has been replaced by  successive copies of  plus an edge.

Suppose that there is a subgraph  in  that is isomorphic to a graph of . Since  is -subgraph-free, and since  is obtained from  by removing one edge, must also be -subgraph-free. Therefore  is not fully contained in a copy of . Since  is -connected, this implies that there is a cycle in  that is not fully contained in a copy of . By definition, this cycle has length at most , a contradiction.

Suppose the graph  has a -colouring . For each copy of , the vertices corresponding to  and  must be coloured the same. For all edges  with in , there is a vertex in  coloured the same as  in  that is adjacent to  in , so . Therefore  restricted to  is a -colouring of .

On the other hand, suppose the graph  has a -colouring . We can extend this -colouring to  by doing the following: for all edges  with in , colour every  in such a way that the vertex corresponding to  and the vertex corresponding to  have colour . This leads to a -colouring of .

Case 2.  contains two vertices  and  with .
As  is connected, we assume without loss of generality that  and  are adjacent; let .

We reduce from the problem of -List Colouring restricted to planar graphs of girth at least in which each vertex has list , , or . This problem is NP-complete by Theorem 2.1. Let a graph  and list assignment  be an instance of this problem. We will construct a planar -subgraph-free graph  with a -regular list assignment  such that  has a colouring that respects  if and only if  has a colouring that respects .

We define . Then, by minimality, has at least one colouring respecting , and moreover, for any colouring of  that respects , and  are coloured alike (otherwise we would have a colouring of  that respects ). Let  be the set of possible colours that can be used on  and  in colourings of  that respect  and let . As , we have . Let us assume, without loss of generality, that and that .

We say that a vertex  in  is a bivertex or trivertex if  is  or , respectively. We construct a planar -free graph .

First suppose that . For each bivertex  in , we do as follows. We add a copy of  to  which we label . The vertex in  corresponding to  is labelled  and we set . We add the edge . This completes the construction of  when .

Now suppose that . Let if  is even and if  is odd (so  is even in both cases). For each bivertex  in , we add  copies of  to  which we label , . The vertex in  corresponding to  is labelled . Let . Add edges such that, for each bivertex  in , the union of  and  induces a cycle on vertices. This completes the construction of  when .

Note that  is planar, since it is made of planar graphs ( and copies of ) connected in a way that does not obstruct planarity. Suppose that there is a subgraph  in  that is isomorphic to a graph of . Since  is -subgraph-free, and since  is obtained from  by removing one edge, is also -subgraph-free. Therefore for all , is not fully contained in . Since  is -connected and since for all , only one vertex of  has a neighbour outside of , we find that  has at most one vertex in each . This means that  is a subgraph of , which contradicts the fact that  has girth at least . Therefore  is -subgraph-free.

Now we define a list assignment . We give the vertices of each copy of  the same lists as their corresponding vertices in , and for each bivertex  in , we define , and for each trivertex  in , we define . This gives us the -regular list assignment  of .

The graph has a colouring that respects (the restriction of)  and we notice that in such a colouring each copy of  must be coloured in such a way that, for each bivertex  in , the  vertices of  that are adjacent to  are coloured with the  colours of . So one of the neighbours of  in  must be coloured . Thus the problem of finding a colouring of  that respects  is equivalent to the problem of finding a colouring of  that respects . The proof is complete.∎

2.3 The Proof of Theorem 1.7

Theorem 1.7 is not quite implied by Theorem 1.6. However, we can adapt the proof of Theorem 1.6 to prove Theorem 1.7, which we restate below.

Theorem 1.7 (restated). -Regular List Colouring is NP-complete for planar graphs with no -cycles, no -cycles and no intersecting triangles.

Proof

Let  be the graph on five vertices with two triangles sharing exactly one vertex (this graph is known as the butterfly). Consider the previous proof with . Note that the only problem is that  is not -connected.

Consider the example in [24] of a graph with no -cycle, no -cycle and no intersecting triangles that is not -choosable. Let  be this example, with  a -regular list assignment such that there is no colouring of  respecting . We remove edges and vertices from  until any further removal would give a graph with a colouring respecting . This leads to a connected graph . There are no four vertices in  with the same list inducing a connected subgraph, so there are no such four vertices in . Therefore in  there is an edge  such that (otherwise  would have at most three vertices, and thus would be -choosable).

Therefore we can skip Case  and directly adapt the proof in Case . Note that the only thing to prove is that in this case  does not contain a subgraph isomorphic to . Suppose  is such a subgraph. Since  is -subgraph-free, cannot be fully contained in  for any . Since no vertex is in two different  and no vertex of  has two adjacent neighbours outside , this implies that there is a triangle in , which is impossible since  has girth at least .∎

2.4 The Proof of Theorem 1.8

The proof is obtained by a modification of the NP-hardness construction of -List Colouring for -free graphs from [17]. Recall that we included this result in our paper to illustrate that -Choosability and -Regular List Colouring can have different complexities when restricted to special graph classes. Indeed, since Choosability is polynomial-time solvable on -free graphs [15], Theorem 1.8 shows that -Choosability may even be easier than -Regular List Colouring.

Theorem 1.8 (restated). -Regular List Colouring is NP-complete for -free graphs.

Proof

The problem is readily seen to belong to NP. Golovach et al. [15] showed that -List Colouring is NP-complete for -free graphs in which every vertex has a list of size  or . Let  be such an instance. We add three new vertices to . We make adjacent to each other and to each original vertex of . This results in a -free graph . We take three new colours and set . This forces colour  to be used to colour one of or . Then all that remains is to add colour  to the list of every vertex of  that has a list of size .∎

2.5 The Proof of Theorem 1.12

We need some additional results. We begin with a theorem of Bonomo et al.

Theorem 2.2 ([3])

List Colouring is polynomial-time solvable on block graphs.

By generalizing their proof we extend this result to classes of graphs where List Colouring is polynomial-time solvable on the blocks of graphs in the class.

Theorem 2.3

Let  be a class of graphs and let  be the class of graphs that appear as blocks of graphs in . If List Colouring is polynomial-time solvable on  then it is polynomial-time solvable on .

Proof

Let  be a graph in  that, together with a list assignment , forms an instance of List Colouring. We may assume  is connected. If then consider a cut-vertex  in a leaf-block  (such  and  exist). Let  be the restriction of  to . For each colour , we do as follows. We remove  from the list of every neighbour of  in  and check whether  admits a colouring that respects . Note that we can do this in polynomial time, as  is in . If so then we put  in a set  for vertex ; otherwise we do not do this. If, after considering each colour in , we find that then we return no. Otherwise we define a new list assignment  for the subgraph  of  induced by by setting and if . Note that  has a colouring that respects  if and only if  has a colouring that respects . We continue with the pair . We do this exhaustively until we obtain in polynomial time a graph in . Since List Colouring is polynomial-time solvable in , this completes the proof.∎

Theorem 2.3, combined with Theorem 1.9, leads to the following generalization of Theorem 2.2.

Corollary 9

List Colouring is polynomial-time solvable on Gallai trees.

We also state the following theorem, proved independently by Borodin and Erdős et al.

Theorem 2.4 ([4, 5, 13])

Let be a connected graph with a list assignment  such that for all . If  is not a Gallai tree then  has a colouring respecting .

We now restate and prove Theorem 1.12.

Theorem 1.12 (restated). Let  be a positive integer. Then -Precolouring Extension is polynomial-time solvable for graphs of maximum degree at most .

Proof

Let , together with a -precolouring  defined on a subset , be an instance of -Precolouring Extension. We let  be the subgraph of  induced by . For each , we set . Observe that  has a -colouring extending  if and only if  has a colouring that respects . Hence we may consider  instead.

Note that, for every vertex , the colours that are removed from to obtain  are exactly the colours of those neighbours of  in  that are not in . This implies, together with the assumption that  has maximum degree at most , that every has at most  neighbours in .

We now apply the following procedure on  exhaustively. If a vertex  has fewer than  neighbours then remove  from . In the end we obtain a graph  with the property that for all . Moreover, has a colouring that respects  if and only if  has a colouring that respects the restriction of  to . Hence we may consider  instead. We consider every connected component  of  in turn. If  is a Gallai tree then we apply Corollary 9. Otherwise we apply Theorem 2.4.∎

3 Conclusions

As well as filling the complexity gaps of a number of colouring problems for graphs with bounded maximum degree, we have given several dichotomies for the -Regular List Colouring problem restricted to subclasses of planar graphs. In particular we showed NP-hardness of the cases and restricted to planar -subgraph-free graphs for several sets  of -connected planar graphs. Our method implies that for such sets  it suffices to find a counterexample to -choosability or to -choosability, respectively. It is a natural question whether we can determine the complexity of -Regular List Colouring and -Regular List Colouring for any class of planar -subgraph-free graphs. However, we point out that even when restricting  to be a finite set of -connected planar graphs, this would be very hard (and beyond the scope of this paper) as it would require solving several long-standing conjectures in the literature. For example, when , Montassier [23] conjectured that every planar -subgraph-free graph is -choosable.

A drawback of our method is that we need the set of graphs  to be -connected. If we forbid a set  of graphs that are not -connected, the distinction between polynomial-time solvable and NP-complete cases is not clear, and both cases may occur even if we forbid only one graph. We illustrate this below with an example.

Example. Let  contain only the star  for some . Note that -subgraph-free graphs are exactly those graphs that have maximum degree at most . Hence, if , then -Regular List Colouring is polynomial-time solvable due to Theorem 1.9. However, there exist larger values of  for which the problem is NP-complete. In order to see this we adapt the proof of Theorem 1.6. The hardness reductions in this proof multiply the maximum degree of our instances by some constant  that is at most the maximum degree of the no-instance . By Theorems 1.2 and 1.4, the problems we reduce from are NP-complete even for graphs with maximum degree at most . Hence, we have proven the following: if  is a finite set of -connected planar graphs and  is a non--choosable planar -subgraph-free graph with maximum degree , then -Regular List Colouring is NP-complete on planar -subgraph-free graphs with maximum degree at most . We can take to deduce that -Regular List Colouring is NP-complete on planar -subgraph-free graphs.

Acknowledgements. We thank Steven Kelk for helpful comments on an earlier version of this paper.

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